Answer is: there is 1,77·10²³ molecules of CH₂Cl₂.
m(CH₂Cl₂) = 25 g.
n(CH₂Cl₂) = m(CH₂Cl₂) ÷ M(CH₂Cl₂).
n(CH₂Cl₂) = 25 g ÷ 85 g/mol.
n(CH₂Cl₂) = 0,294 mol.
N(CH₂Cl₂) = n(CH₂Cl₂) · Na.
N(CH₂Cl₂) = 0,294 mol · 6,023 1/mol.
N(CH₂Cl₂) = 1,77·10²³.
n - amount of substance.
Na - Avogadro number.
Question 1 :
V1/T1 = V2/T2
3.0L/273K = V2/373K
To get the value of Z, cross multiply
3.0L x 373K = 273K x V2
1119 = 273V2
Divide both sides by 273
1119/273 = 273V2/273
4.10L = V2
The new volume is 4.10 liters
Question 2 :
P1/T1 = P2 /T2
P1 = 880 kPA= 880 *10^3 Pa
T1 = 250 K
T2 = 303 K
P2 =?
Substituting for P2
P2 = P1 T2/ T1
P2 = 880 kPa * 303 / 250
P2 = 266,640 kPa/ 250
P2 = 1066.56 kPa.
The new pressure of the gas is 1066.56 kPa
Question 3 :
Given that:
Volume of gas V = 4.80L
(since 1 liter = 1dm3
4.80L = 4.80dm3)
Temperature T = 62°C
Convert Celsius to Kelvin
(62°C + 273 = 335K)
Pressure P = 2.9 atm
Number of moles of gas N = ?
Apply ideal gas equation
pV = nRT
2.9atm x 4.8dm3 = n x (0.0082 atm dm3 K-1 mol-1 x 335K)
13.92 atm dm3 = nx 2.747 atm dm3 mol-1
n = 13.92/2.747
n = 5.08 moles
There are 5.08 moles of gas contained in the sample
Question 4 :
Volume of gas V = 3.47L
(since 1 liter = 1dm3
3.47L = 3.47dm3)
Temperature T = 85.0°C
Convert Celsius to Kelvin
(85.0°C + 273 = 358K)
Pressure P = ?
Number of moles of gas N = 0.100 mole
Apply ideal gas equation
pV = nRT
p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)
p x 3.47dm3 = 0.29 atm dm3
p = (0.29 atm dm3 / 3.47 dm3)
p = 0.085 atm
If 1 atm = 760 mm Hg
0.085atm = 0.085 x 760
= 64.6 mm Hg
The pressure of the gas is 64.6 mm hg
Answer:
c.boron-11
Explanation:
The atomic mass of boron is 10.81 u.
And 10.81 u is a lot closer to 11u than it is to 10u, so there must be more of boron-11.
To convince you fully, we can also do a simple calculation to find the exact proportion of boron-11 using the following formula:
(10u)(x)+(11u)(1−x)100%=10.81u
Where u is the unit for atomic mass and x is the proportion of boron-10 out of the total boron abundance which is 100%.
Solving for x we get:
11u−ux=10.81u
0.19u=ux
x=0.19
1−x=0.81
And thus the abundance of boron-11 is roughly 81%.
one mole of P weights about 31 grams
in one mole there are 6.022*10^23 atoms
we use the rule of threes
6.022*10^23atoms......weight..........31 grams
3.45*10^23 atoms.........weight...........x grams
x=(3.45*10^23*31)/6.022*10^23
x=106.95/6.022=<u><em>17.76 grams</em></u>
